How to change the display_channel with python?
Make sure you have the correct context.

The context area needs to be a video sequencer. Quick console code, find the VSE and set the property from py console.

&gt;&gt;&gt; for i, a in enumerate(C.screen.areas):
...     i, a.type
...     
(0, 'INFO')
(1, 'PROPERTIES')
(2, 'CONSOLE')
(3, 'VIEW_3D')
(4, 'SEQUENCE_EDITOR')

&gt;&gt;&gt; C.screen.areas[4].spaces.active.display_channel
0


If your script is invoked via a button or menu in the VSE area use

space = context.space_data


or 

space = context.area.spaces.active


If it is not a VSE area / space then you will get an error (unless another space has a display_channel property.

Each area can have at least one up to  a space for each of the editor types. (SEQUENCER, CONSOLE, etc)  The one you are seeing is the active one. area.spaces.active Others you have had active at some stage are stored in area.spaces. 

EDIT:

Find all preview VSE's in screen, and set display channel.

previews = [a.spaces.active for a in context.screen.areas
        if a.type == 'SEQUENCE_EDITOR'
        and a.spaces.active.view_type == 'PREVIEW']

for s in previews:
    s.display_channel = 3


As for bpy.types.SpaceSequenceEditor(display_channel = 5) what is the expectation here?. Instancing from bpy.types classes is very rare, and is mostly done via a new or add method.